Question

The total energy radiated from a black body source is collected for $1min$ and is used to heat a quantity of water. The temperature of water is found to increase from $20℃$ to $20.5℃$ . If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $20℃$ , the temperature of water will be

• A. $21^{\circ }C$
• B. $22{}^{\circ }C$
• C. $24{}^{\circ }C$
• D. $28{}^{\circ }C$

Answer 1

Answer: (D)

From Stefan's law of radiation,
$E\propto T^4\Rightarrow \frac{E_1}{E_2}=\frac{T_1^4}{T_2^4}$
Given, $T_1=T,T_2=2T$
$\therefore$ $\frac{E_1}{E_2}=\frac{{\left(T\right)}^4}{{\left(2T\right)}^4}=\frac{1}{2^4}=\frac{1}{16}$
$E_2=16E_1$
Heat taken by water from radiation
$E=mc\Delta \theta$
Where c is specific heat, $\Delta \theta$ the change in temperature and $m$ the mass.
$\therefore$ $E=m\times 1\times (20.5-20)$
$E=m\times 0.5$ ...(i)
When energy supplied is 16 times the previous one, then let the temperature rise to $\theta \prime$
$\therefore 16E=m\times 1\times ({\left(\theta \right)}^{\prime }-20)$ ...(ii)
Dividing Eq. (i) by (ii), we get
$\frac{1}{16}=\frac{0.5}{{\theta }^{\prime }-20}$
$\Rightarrow {\theta }^{\prime }-20=16\times 0.5=8$
$\Rightarrow {\theta }^{\prime }=20+8=28℃$